Prove that lim_(x→0) ((1−cos(x)cos(x/2)cos(x/3)...)/x^2 )=(π^2 /(12))


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Question Number 58309 by Smail last updated on 21/Apr/19

$P r o v e t h a t$ $\underset{x \to 0}{l i m} \frac{1 - c o s (x) c o s (x / 2) c o s (x / 3) . . .}{x^{2}} = \frac{π^{2}}{12}$
Commented by maxmathsup by imad last updated on 21/Apr/19
$let use hospital theorem first we have by chang.x=6t lim_(x→0) ((1−cos(x)cos((x/2))cos((x/3)))/x^2 ) =lim_(t→0) ((1−cos(6t) cos(3t)cos(2t))/(36 t^2 )) but we have cos(6t)cos(3t) =(1/2) {cos(9t)+cos(3t)} ⇒ cos(6t)cos(3t)cos(2t) =(1/2){cos(9t)cos(2t) +cos(3t)cos(2t)} =(1/4){cos(11t)+cos(7t) +cos(5t) +cos(t)} let f(t) =1−(1/4){cos(t) +cos(5t) +cos(7t) +cos(11t)} and g(t)=36x^2 ⇒ f^′ (t) =(1/4){sin(t)+5sin(5t)+7sin(7t)+11sin(11t)} ⇒ f^(′′) (t) =(1/4){cost +25 cos(t) +49 cos(7t) +121 cos(11t)} ⇒ lim_(t→0) f^(′′) (t) =(1/4){1+25 +49 +121} =(1/4){26 +49 +121} =(1/4){121 +75} =(1/4)(196) =((98)/2) =49 also we have g^′ (t) =72 x and g^(′′) (x) =72 ⇒ lim_(x→0) ((f^(′′) (t))/(g^(′′) (t))) =((49)/(72)) so there is a error in the question ...$
$l e t u s e h o s p i t a l t h e o r e m f i r s t w e h a v e b y c h a n g . x = 6 t$ ${l i m}_{x \to 0} \frac{1 - c o s (x) c o s (\frac{x}{2}) c o s (\frac{x}{3})}{x^{2}} = {l i m}_{t \to 0} \frac{1 - c o s (6 t) c o s (3 t) c o s (2 t)}{36 t^{2}}$ $b u t w e h a v e c o s (6 t) c o s (3 t) = \frac{1}{2} {c o s (9 t) + c o s (3 t)} \Rightarrow$ $c o s (6 t) c o s (3 t) c o s (2 t) = \frac{1}{2} {c o s (9 t) c o s (2 t) + c o s (3 t) c o s (2 t)}$ $= \frac{1}{4} {c o s (11 t) + c o s (7 t) + c o s (5 t) + c o s (t)} l e t$ $f (t) = 1 - \frac{1}{4} {c o s (t) + c o s (5 t) + c o s (7 t) + c o s (11 t)} a n d g (t) = 36 x^{2} \Rightarrow$ $f^{^{'}} (t) = \frac{1}{4} {s i n (t) + 5 s i n (5 t) + 7 s i n (7 t) + 11 s i n (11 t)} \Rightarrow$ $f^{^{″}} (t) = \frac{1}{4} {c o s t + 25 c o s (t) + 49 c o s (7 t) + 121 c o s (11 t)} \Rightarrow$ ${l i m}_{t \to 0} f^{^{″}} (t) = \frac{1}{4} {1 + 25 + 49 + 121} = \frac{1}{4} {26 + 49 + 121}$ $= \frac{1}{4} {121 + 75} = \frac{1}{4} (196) = \frac{98}{2} = 49 a l s o w e h a v e g^{^{'}} (t) = 72 x a n d$ $g^{^{″}} (x) = 72 \Rightarrow {l i m}_{x \to 0} \frac{f^{^{″}} (t)}{g^{^{″}} (t)} = \frac{49}{72} s o t h e r e i s a e r r o r i n t h e q u e s t i o n . . .$
Commented by mr W last updated on 21/Apr/19

$q u e s t i o n i s c o r r e c t s i r .$ ${i t}^{'} s \lim_{x \to 0} \frac{1 - \cos (x) \cos (x / 2) \cos (x / 3) . . . . .}{x^{2}}$ $n o t \lim_{x \to 0} \frac{1 - \cos (x) \cos (x / 2) \cos (x / 3)}{x^{2}}$
Commented by Mr X pcx last updated on 21/Apr/19

$i h a v n t s e e n t h e p o i n t b u t n r v e r m i n d$ $i h a v e t r e a t e d a s p e c i a l c a s e . . .$
	Answered by tanmay last updated on 21/Apr/19

	$c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + . . .$ $h e r e w e t a k e c o s x \approx 1 - \frac{x^{2}}{2} o t h e r t e r m s i g n o r e d$ $b e c a u s e t h o s e t e r m s c o n t a i n x^{r} w h e n r > 2$ $f (x) = c o s x c o s (\frac{x}{2}) c o s (\frac{x}{3}) c o s (\frac{x}{4}) . . .$ $f (x) = (1 - \frac{x^{2}}{2 \times 1^{2}}) (1 - \frac{x^{2}}{2 \times 2^{2}}) (1 - \frac{x^{2}}{2 \times 3^{2}}) . . .$ $f (x) = \underset{r = 1}{\prod^{\infty}} (1 - \frac{x^{2}}{2 \times r^{2}})$ $f (x) = 1 - \frac{x^{2}}{2} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .) + o t h e r s t e r m s i g n o r e d$ $s o v a l u e o f$ $l i \underset{x \to 0}{m} \frac{1 - f (x)}{x^{2}}$ $= l i \underset{x \to 0}{m} \frac{\frac{x^{2}}{2} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .)}{x^{2}}$ $= \frac{1}{2} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .) = \frac{1}{2} \times \frac{π^{2}}{6} = \frac{π^{2}}{12}$
	Commented by Tawa1 last updated on 21/Apr/19

	$Wow great sir .$ $But how is, \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . = \frac{π^{2}}{6}$
	Commented by tanmay last updated on 21/Apr/19

	Commented by tanmay last updated on 21/Apr/19

	$p l s c h e c k s r l n o 19.19$
	Commented by Smail last updated on 21/Apr/19

	$t h a n k s$
	Commented by tanmay last updated on 21/Apr/19

	$t h e q u e s t i o n c r e a t e d r i p p l e s i n t h o u g h t p r o c e s s$ $l a t e r i f o u n d t h e l i g h t t o r e a c h g o a l . . .$ $i h a v e s o l v e d b y e x p a n s i o n m e t h o d . .$
	Commented by Tawa1 last updated on 21/Apr/19

	$How can we prove that \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . \infty = \frac{π^{2}}{6}$
	Commented by peter frank last updated on 21/Apr/19

	$t h a n k y o u$
	Commented by maxmathsup by imad last updated on 22/Apr/19

	$s i r T a w a t a k e a l o o k a t t h e p l a t e f o r m t h i s r e s u l t i s p r o v e d b y d i f f e r e n t$ $m e t h o d s . . .$
	Commented by Tawa1 last updated on 22/Apr/19

	$I {don}^{'} t know anyone sir$
	Commented by Smail last updated on 22/Apr/19

	$C h e c k a c h a n n e l c a l l e d o n Y o u t u b e c a l l e d$ $3 B l u e 1 B r o w n . T h e y u s e d a b r i l l i a n t$ $w a y t o p r o o f t h a t i d e n t i t y .$
	Answered by mr W last updated on 21/Apr/19
	$let f(x)=cos (x) cos (x/2) cos (x/3) ... lim_(x→0) f(x)=1 ln f(x)=cos (x)+cos (x/2)+cos (x/3) ... ((f′(x))/(f(x)))=−sin (x)−(1/2)sin (x/2)−(1/3)sin (x/3)−.... f′(x)=−f(x)[sin (x)+(1/2)sin (x/2)+(1/3)sin (x/3)+....] lim_(x→0) f′(x)=0 f′′(x)=−f′(x)[sin (x)+(1/2)sin (x/2)+(1/3)sin (x/3)+....]−f(x)[cos x+(1/2^2 )cos (x/2)+(1/3^2 )cos (x/3)+...] =f(x){[sin (x)+(1/2)sin (x/2)+(1/3)sin (x/3)+....]^2 −[cos x+(1/2^2 )cos (x/2)+(1/3^2 )cos (x/3)+...]} lim_(x→0) f′′(x)=1×{[0+0+0+...]^2 −[1+(1/2^2 )+(1/3^2 )+...]}=−(1+(1/2^2 )+(1/3^2 )+...)=−(π^2 /6) lim_(x→0) ((1−cos(x)cos(x/2)cos(x/3)...)/x^2 ) =lim_(x→0) ((1−f(x))/x^2 ) (=(0/0)) =lim_(x→0) ((−f′(x))/(2x)) (=(0/0)) =lim_(x→0) ((−f′′(x))/2) =(1/2)(1+(1/2^2 )+(1/3^2 )+...) =(1/2)×(π^2 /6) =(π^2 /(12))$
	$l e t f (x) = \cos (x) \cos (x / 2) \cos (x / 3) . . .$ $\lim_{x \to 0} f (x) = 1$ $\ln f (x) = \cos (x) + \cos (x / 2) + \cos (x / 3) . . .$ $\frac{f^{'} (x)}{f (x)} = - \sin (x) - \frac{1}{2} \sin \frac{x}{2} - \frac{1}{3} \sin \frac{x}{3} - . . . .$ $f^{'} (x) = - f (x) [\sin (x) + \frac{1}{2} \sin \frac{x}{2} + \frac{1}{3} \sin \frac{x}{3} + . . . .]$ $\lim_{x \to 0} f^{'} (x) = 0$ $f^{″} (x) = - f^{'} (x) [\sin (x) + \frac{1}{2} \sin \frac{x}{2} + \frac{1}{3} \sin \frac{x}{3} + . . . .] - f (x) [\cos x + \frac{1}{2^{2}} \cos \frac{x}{2} + \frac{1}{3^{2}} \cos \frac{x}{3} + . . .]$ $= f (x) {{[\sin (x) + \frac{1}{2} \sin \frac{x}{2} + \frac{1}{3} \sin \frac{x}{3} + . . . .]}^{2} - [\cos x + \frac{1}{2^{2}} \cos \frac{x}{2} + \frac{1}{3^{2}} \cos \frac{x}{3} + . . .]}$ $\lim_{x \to 0} f^{″} (x) = 1 \times {{[0 + 0 + 0 + . . .]}^{2} - [1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .]} = - (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .) = - \frac{π^{2}}{6}$ $\underset{x \to 0}{l i m} \frac{1 - c o s (x) c o s (x / 2) c o s (x / 3) . . .}{x^{2}}$ $= \underset{x \to 0}{l i m} \frac{1 - f (x)}{x^{2}} (= \frac{0}{0})$ $= \underset{x \to 0}{l i m} \frac{- f^{'} (x)}{2 x} (= \frac{0}{0})$ $= \underset{x \to 0}{l i m} \frac{- f^{″} (x)}{2}$ $= \frac{1}{2} (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .)$ $= \frac{1}{2} \times \frac{π^{2}}{6}$ $= \frac{π^{2}}{12}$
	Commented by tanmay last updated on 21/Apr/19

	$b a h d a r u n s i r . . . L H o s p i t a l r u l e e x c e l l e n t . . .$
	Commented by Mr X pcx last updated on 21/Apr/19

	$s i r m r w h a v e p l a y e d a k r i k e t m a t c h$ $w i t h t h i s l i m i t . . .$
	Commented by tanmay last updated on 29/Apr/19

	$f (x) = c o s x c o s (\frac{x}{2}) c o s (\frac{x}{3}) . . .$ $l n f (x) = l n (c o s x) + l n c o s (\frac{x}{2}) + l n c o s (\frac{x}{3}) + . . .$ $\frac{f^{^{'}} (x)}{f (x)} = - [t a n x + \frac{1}{2} t a n (\frac{x}{2}) + \frac{1}{3} t a n (\frac{x}{3}) + . . .]$ $s i r p l s c h e c k . . .$
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